Well if all the coefficients are non-zero, then the degree of $p_n(x)$ is $n$, whilst the degree of $\prod_{i=1}^{n-1}p_i(x)$ is $(n-1)!$. For the difference of these two to be constant, we need $n=(n-1)!$, which holds only for $n=1$. Therefore there are no solutions for $n\ge2$. Right?

Diarmuid wrote: whilst the degree of $\prod_{i=1}^{n-1}p_i(x)$ is $(n-1)!$ No, the degree of $\prod_{i=1}^{n-1}p_i(x)$ is $1+2+\ldots +(n-1)$, which implies $n=3$. A short computation shows that $P(x)=2a^3x^3+2a^2x^2+ax+\frac 12$ for some $a \in \mathbb{R}^*$.

Oops... Sorry, I make a lot of mistakes when I'm tired...